ANSWER
a) The main point is to identify that the one-year change in longer-term yields is complementary on the left hand
side, and the multi-year return on long term bonds held to one year before maturity is complementary on the right
hand side. The graph is good.
Time
Price
Higher 1 year
rate in year 1
-2
= lower return of 2
year bond in 0- 1
Time
=Higher 1 year
rate in 3-4
Lower return of 4
year bond from 0-3
Time
Lower return of 4
year bond from 0-1
= Higher 3 year
rate in 1-4
‘safe’ 1 year return
b) To really get this right you have to identify the long/short positions. Left side:
(3)
−
(1)
=
(2)
−
(3)
−
(1)
=
−
³
(2)
+1
−
(2)
´
+
³
(2)
+1
−
(3)
´
−
(1)
=
2
³
(2)
+1
−
(2)
´
+
(2)
+1
73
If you run a regression of left and right side on
(3)
−
(1)
you
fi
nd that 1 (left) = (right) coe
ffi
cient of
2
³
(2)
+1
−
(2)
´
on
(3)
−
(1)
plus
the coe
ffi
cient of
(2)
+1
on
(3)
−
(1)
.
I did not ask for the right side, but here it is.
(3)
−
(1)
=
(2)
−
(3)
−
(1)
=
−
(1)
+2
+
(2)
+
(1)
+2
−
(3)
−
(1)
=
(1)
+2
−
(3)
+
(2)
+
³
(1)
+2
−
(1)
´
=
³
(3
→
1)
→
+2
−
(2
→
0)
→
+2
´
+
³
(1)
+2
−
(1)
´
Corresponding to the two-year change in
(1)
then is the excess return for buying a 3 year bond and holding for two
years, over the return for buying a 2 year bond and holding for 2 years.
12) (15) You form a model of the term structure of interest rates by supposing the one-year rate is an AR(1),
(1)
+1
−
=
(
(1)
−
) +
+!
and that the expectations hypothesis holds.
If your model is right, what should the eigenvalue decomposition of
forward rates look like? Speci
fi
cally, if you took data as predicted from your model,
≡
⎡
⎢
⎣
(1)
(2)
(3)
⎤
⎥
⎦
and performed
Λ
0
=
(
(
0
))
, what would the
and
Λ
look like? Note: you do not have to give the exact
values of
and
Λ
. It is enough to answer that columns of
have a speci
fi
c pattern, show where any zeros are, and
say what if any parts of
or
Λ
are arbitrary and don’t matter.
ANSWER: Forward rates should follow a one-factor model.
(2)
=
(
(1)
+1
) =
+
(
(1)
−
)
(3)
=
(
(1)
+2
) =
+
2
(
(1)
−
)
(
)
=
(
(1)
+
−
1
) =
+
−
1
(
(1)
−
)
thus
Λ
will have only one non-zero element
Λ
=
⎡
⎣
1
0
0
0
0
0
0
0
0
⎤
⎦
and
will look like
=
⎡
⎣
1
|
|
2
3
2
|
|
⎤
⎦
where
is an arbitrary constant —
the pattern is
1
2
. Since they multiply zeros,
the second and third columns of
are irrelevant
.
